Question

It is not practicable to measure the intercept directly at 1/T=0, so calculate ln A from the measured slope and the coordinates of any point on the graph. Compare your values of E and A with those obtained by previous students, namely E = 52 kJ mol^-1 and A = ~10⁷ dm³ mol^-1 s^-1

For E = 45.3953 kJ mol^-1

^equation: y = -5460.1x + 10.272

At 273K, 301K and 308K, I calculate the A, but it's ~3 x 10^-3 dm³ mol^-1 s^-1

It is not practicable to measure the intercept directly at 1/T-0, so calculate In A from the measured slope and the coordinates of any point on the graph. Compare your values of E and A with those obtained by previous students, namely E 52 kJ mol 1 and A 10 dm mol-1 s For E 453953 KJ mol toy-5460.1x+10.272 At 273K, 301K and 308K, I calculate the A, but it's -3 x 10 dm mol 1 s 0.0032 0.0033 0.0034 0.0035 o.0036 0.0037 01x-10.22.. R0.8167 -10 12 1/T

Answer 1

Given dato Qcc to ahemu theB tntes ce Given slope op-tbe Ore om te above dta A36dm mois K6.6667 XO Promo a. Puttin th above vacue -6.6667 x 10,485-r108

A 15 85 mmo's ptat 08k9-6155 45:39 53X10 4.6155 x to e 308x8914